2.1 Dielectric slab waveguide a Consider the rays 1 and 2 in Figure 2.3. Derive the waveguide condition.

b Consider the two rays 1 and 2 in Figure 2.4. Show that the phase difference when they meet at C at a distance y above the guide center is

Φm = k12(a − y)cosθm − φm

c Using the waveguide condition show that

Φm = Φm(y) = mπ −ya

(mπ + φm )

Solution

n2

n2

z2a

y

A

1

2 1

B

θθ

θA′

B′

Cπ−2θ 2θ−π/2

k1

Ex

n1

Figure 2.3 Two arbitrary waves 1and 2 that are initially in phase mustremain in phase after reflections.Otherwise the two will interferedestructively and cancel each other.

n2

z

a y

A

1

2

θθ

A′

C

kE

x

y

a−y

Guidecenter

π−2θ

Figure 2.4 Interference of waves such as 1and 2 leads to a standing wave patternalong the y-direction which propagatesalong z.

From the geometry we have the following:

(a − y)/AC = cosθ

and Α′C/AC = cos(π − 2θ)

The phase difference between the rays meeting at C is

Φ = kAC − φ − kA′C = k1AC − k1ACcos(π − 2θ) − φ

= k1AC[1 − cos(π − 2θ)] − φ = k1AC[1 + cos(2θ)] − φ

= k1[(a − y)/cosθ][ 1 + 2cos2θ − 1]− φ

= k1 [(a − y)/cosθ][2cos2θ]− φ

= 2k1(a − y)cosθ− φ

Given, 2π (2a)n1

λ⎡ ⎣

⎤ ⎦ cosθm −φm = mπ

∴ cosθm =λ (mπ + φm )

2πn1(2a)=

mπ +φm

k1(2a)

Then Φm = 2k1(a − y) cosθm −φm = 2k1(a − y)mπ +φm

k1(2a)− φm

∴ Φm = (1 −ya

)(mπ +φm) −φm = mπ −ya

(mπ +φm)

Φm = Φm(y) = mπ −ya

(mπ + φm )

2.7 Dielectric slab waveguide Consider a planar dielectric waveguide with a core thickness 10 μm, n1 = 1.4446, n2 = 1.4440. Calculate the V-number, the mode angle θm for m = 0 (use a graphical solution, if necessary), penetration depth, mode field distance (MFD = 2α + 2δ), for light wavelengths of 1.0 μm and 1.5 μm. What is your conclusion? Compare your MFD calculation with 2wo = 2a(V+1)/V.

Solution

λ = 1 μm, n1 = 1.4446, n2 = 1.4440, a = 5 μm. Apply

V =2πa

λn1

2 − n22( )1/ 2

to obtain V = 1.3079

Solve the waveguide condition

tan ak1 cosθm − mπ2

⎛ ⎝

⎞ ⎠ =

sin2 θm − n2

n1

⎛ ⎝ ⎜ ⎞

⎠ ⎟

2⎡

⎣ ⎢

⎤

⎦ ⎥

1/ 2

cosθm

= f (θm)

graphically as in Example 2.1.1 to find: θc = 88.35° and the mode angle (for m = 0) is θo = 88.85°.

Then use

1

δm

= αm =

2πn2n1

n2

⎛ ⎝ ⎜ ⎞

⎠ ⎟

2

sin2 θm −1⎡

⎣ ⎢

⎤

⎦ ⎥

1/ 2

λ

to calculate the penetration depth:

δ = 1/α = 5.33 μm.

∴ MFD = 2a + 2δ = 20.65 μm

We can also calculate MFD from

MFD = 2a(V+1)/V = 2(5 μm)(1.3079+1)/(1.3079) = 17.6 μm (Difference = 15%)

λ = 5 μm, V = 0.262, single mode. Solve waveguide condition graphically to find that the mode angle is θo = 88.40°.

δ = 1/α = 77.22 μm.

∴ MFD = 2a + 2δ = 164.4 μm.

Compare with MFD = 2a(V+1)/V = 2(5 μm)(0.262 + 1)/(0.262) = 48.2 μm (Very large difference)

Notice that the MFD from 2a(V+1)/V gets worse as V decreases. The reason for using MFD = 2a(V+1)/V, this equation provides a single step calculation of MFD. The calculation of the penetration depth d requires the calculation of the incidence angle θ and φ.

λ = 1.5 μm, V = 0.872, single mode. Solve waveguide condition graphically that the mode angle is θo = 88.72°.

δ = 1/α = 9.08 μm.

∴ MFD = 2a + 2δ = 28.15 μm.

Compare with MFD = 2a(V+1)/V = 2(5 μm)(0.872+1)/(0.872) = 21.5 μm (Difference = 24%)

Notice that the MFD from 2a(V+1)/V gets worse as V decreases. The reason for using MFD = 2a(V+1)/V, this equation provides a single step calculation of MFD. The calculation of the penetration depth δ requires the calculation of the incidence angle θ and φ.

2.8 A multimode fiber Consider a multimode fiber with a core diameter of 100 μm, core refractive index of 1.475 and a cladding refractive index of 1.455 both at 850 nm. Consider operating this fiber at λ = 850 nm.

a Calculate the V-number for the fiber and estimate the number of modes.

b Calculate the wavelength beyond which the fiber becomes single mode.

c Calculate the numerical aperture.

d Calculate the maximum acceptance angle.

e Calculate the modal dispersion Δτ and hence the bit rate × distance product given that rms dispersion σ ≈ 0.29Δτ where Δτ is the full spread.

Solution

Given n1 = 1.475, n2 = 1.455, 2a = 100×10-6 m or a = 50 μm and λ = 0.850 μm. The V-number is,

V =2πaλ

n12 − n2

2( )1/ 2=

2π(50 μm) 1.4752 −1.4552( )1 / 2

(0.850 μm) = 89.47

Number of modes M,

M =V2

2=

89.472

2≈ 4002

The fiber becomes monomode when,

V =2πaλ

n12 − n2

2( )1/ 2< 2.405

or λ >2πa n1

2 − n22( )1/ 2

2.405=

2π(50 μm) 1.4752 − 1.4552( )1/ 2

2.405= 31.6 μm

For wavelengths longer than 31.6 μm, the fiber is a single mode waveguide.

The numerical aperture NA is

NA = n12 − n2

2( )1 / 2= 1.4752 − 1.4552( )1/ 2

= 0.242

If αmax is the maximum acceptance angle, then,

αmax = arcsinNAno

⎛

⎝ ⎜ ⎞

⎠ ⎟ = arcsin(0.242/1) = 14°

Modal dispersion is given by

Δτ intermode

L=

n1 − n2

c=

1.475 −1.4553 ×108 m s−1

= 66.7 ps m-1 or 67.6 ns per km

Given that σ ≈ 0.29Δτ, maximum bit-rate is

BL =0.25Lσ total

≈0.25L

σ intermode

=0.25

(0.29)(66.7 ns km −1)

i.e. BL = 13 Mb s-1 km (only an estimate!)

We neglected material dispersion at this wavelength which would further decrease BL. Material dispersion and modal dispersion must be combined by

σ total2 = σ intermode

2 + σmaterial2

For example, assuming an LED with a spectral rms deviation σλ of about 20 nm, and a Dm ≈ −200 ps km-1 nm-1 (at about 850 nm)we would find

σm = −(200 ps km-1 nm-1)(20 nm)(1 km) ≈ 4000 ps km-1 or 4 ns km-1,

which is substantially smaller than the intermode dispersion and can be neglected.

2.9 A single mode fiber Consider a fiber with a SiO2-13.5%GeO2 core of diameter of 8 μm and refractive index of 1.468 and a cladding refractive index of 1.464 both refractive indices at 1300 nm where the fiber is to be operated using a laser source with a half maximum width of 2 nm.

a Calculate the V-number for the fiber. Is this a single mode fiber?

b Calculate the wavelength below which the fiber becomes multimode.

c Calculate the numerical aperture.

d Calculate the maximum acceptance angle.

e Obtain the material dispersion and waveguide dispersion and hence estimate the bit rate × distance product (B×L) of the fiber.

Solution

a Given n1 = 1.475, n2 = 1.455, 2a = 8×10-6 m or a = 4 μm and λ =1.3 μm. The V-number is,

V =2πaλ

n12 − n2

2( )1/ 2=

2π(4 μm) 1.4682 −1.4642( )1/ 2

(1.3 μm) = 2.094

b Since V < 2.405, this is a single mode fiber. The fiber becomes multimode when

V =2πaλ

n12 − n2

2( )1/ 2> 2.405

or λ <2πa n1

2 − n22( )1/ 2

2.405=

2π(4 μm) 1.4682 −1.4642( )1/ 2

2.405=1.13 μm

For wavelengths shorter than 1.13 μm, the fiber is a multi-mode waveguide.

c The numerical aperture NA is

NA = n12 − n2

2( )1 / 2= 1.4682 − 1.4642( )1/ 2

= 0.108

d If αmax is the maximum acceptance angle, then,

αmax = arcsinNAno

⎛

⎝ ⎜ ⎞

⎠ ⎟ = arcsin(0.108/1) = 6.2°

so that the total acceptance angle is 12.4°.

e At λ =1.3 μm, from the figure, Dm ≈ −7.5 ps km-1 nm-1, Dw ≈ −5 ps km-1 nm-1.

Δτ1/ 2

L= Dm + Dw Δλ1/ 2

= |−7.5−5 ps km-1 nm-1|(2 nm) = 15 ps km-1 + 10 ps km-1

= 0.025 ns km-1

Obviously materials dispersion is 15 ps km-1 and waveguide dispersion is 10 ps km-1 The maximum bit-rate distance product is then

BL ≈0.59LΔτ1 / 2

=0.59

0.025 ns km −1 = 23.6 Gb s-1 km.

Material and waveguide dispersioncoefficients in an optical fiber with acore SiO2-13.5%GeO2 for a = 2.5 to4 μm.

0

–10

10

20

1.2 1.3 1.4 1.5 1.6–20

λ (μm)

Dm

Dw

SiO2-13.5%GeO2

2.53.03.54.0a (μm)

Dispersion coefficient (ps km-1 nm-1)

Figure 2Q9

2.10 A single mode fiber design According to Question 1.3 (Ch.1), the Sellmeier dispersion equation provides n vs. λ for pure SiO2 and SiO2-13.5 mol.%GeO2. The refractive index increases linearly with addition of GeO2 to SiO2 from 0 to 13.5 mol.%. A single mode step index fiber for use at 1300 nm is required to have the following properties: NA = 0.1, core diameter of 9 μm and a core of SiO2-13.5% GeO2. What should be the cladding composition?

Solution Given

2a = 9×10-6 m or a = 4.5 μm. From Ch 1, Question 1.3, the Sellmeier equation is,

nG G G2 1

2

212

22

222

32

2321− =

−+

−+

−

λλ λ

λλ λ

λλ λ

where G1, G2, G3 and λ1, λ2 and λ3 are constants given below where λ1, λ2, λ3 are in μm. Sellmeier constants G1 G2 G3 λ1 λ2 λ3

SiO2-13.5%GeO2 0.711040 0.451885 0.704048 0.0642700 0.129408 9.425478

The fiber is to operate at λ =1.3 μm, thus, using the Sellmeier equation above with the constants in the table we find

n1 = 1.4682

The V-number is,

V =2πaλ

NA =2π(4.5 μm)(1.3 μm)

(0.1) = 2.175

Apply NA = n12 − n2

2( )1 / 2≈ 2n1

2Δ( )1/ 2

or 0.1 ≈ 2(1.4682)2 Δ[ ]1 / 2

to obtain Δ = 0.002320

Apply Δ =n1 − n2

n1

, i.e. 0.0232 =1.4681 − n2

1.4681

Thus, the required cladding refractive index is

n2 = 1.4648

Pure silica has n = 1.4473, SiO2-13.5 mol.%GeO2 has n1 = 1.4682, by linear interpolation the composition corresponding to n2 = 1.4648 is 11.3 mol.% GeO2. Note, the refractive index n(x) of SiO2-x mol.%GeO2, assuming a linear relationship, can be written as

n(x) = n(0) 1−x

13.5⎛ ⎝

⎞ ⎠ + n(13.5)

x13.5

where n(0) = 1.4473; n(13.5) = n1 = 1.4682. Substituting n(x) = 1.4648 gives x = 11.3 .

2.11 Material dispersion If Ng1 is the group refractive index of the core material of a step fiber, then the propagation time (group delay time) of the fundamental mode is

τ =

Lvg

=LNg1

c

Since Ng will depend on the wavelength, show that the material dispersion coefficient Dm is given approximately by

Dm =dτ

Ldλ≈

λc

d2ndλ2

Using the Sellmeier equation in Question 1.3 in Chapter 1 evaluate material dispersion at λ = 1.55 μm for pure silica (SiO2) and SiO2-13.5%GeO2 glass.

Solution From Ch. 1 we know that

Ng ≈ n − λ

dndλ

Differentiate τ with respect to wavelength λ using the above relationship between Ng and n.

τ =

Lvg

=LNg1

c

∴

dτdλ

=Lc

dNg1

dλ≈

Lc

dndλ

− λd2ndλ2 −

dndλ

⎡ ⎣ ⎢

⎤ ⎦ ⎥ = −

Lc

λd 2ndλ2

Thus, Dm =dτ

Ldλ≈

λc

d 2ndλ2 (1)

From Ch. 1 we know that the Sellmeier equation is

nG G G2 1

2

212

22

222

32

2321− =

−+

−+

−

λλ λ

λλ λ

λλ λ

(2)

where G1, G2, G3 and λ1, λ2 and λ3 are constants; given in Table 2Q11 (called Sellmeier coefficients) that are determined by fitting this expression to the experimental data.

We can substitute Eq. (2) in Eq. (1) to obtain Dm and plot Dm vs. λ as shown in Figure 2Q11. The substitution, differentiation and the plot were done on Mathview but almost any other math-software package can do the same. Thus,

At λ = 1.55 μm, Dm = −14 ps km-1 nm-1

Table 2Q11 The Sellmeier coefficients for SiO2-13.5%GeO2.

The λ1, λ2, λ3 are in μm. G1 G2 G3 λ1 λ2 λ3

SiO2-13.5%GeO2 0.711040 0.451885 0.704048 0.0642700 0.129408 9.425478

Wavelength (μm)

Dm (ps km-1 nm-1)

15 ps km-1 nm-1

1.55 μm Figure 2Q11

2.12 Waveguide dispersion Waveguide dispersion arises as a result of the dependence of the propagation constant on the V-number which depends on the wavelength. It is present even when the refractive index is constant; no material dispersion. Let us suppose that n1 and n2 are wavelength (or k) independent. Suppose that β is the propagation constant of mode lm and k = 2π/λ where λ is the free space wavelength. Then the normalized propagation constant b and propagation constant are related by (see Example 2.3.4)

β = n2k[1 + bΔ] (1)

The group velocity is defined and given by

vg =

dωdβ

= cdkdβ

Show that the propagation time, or the group delay time, τ of the mode is

τ =

Lvg

=Ln2

c+

Ln2 Δc

d(kb)dk

(2)

Given the definition of V,

V = ka[n12 − n2

2 ]1 / 2 ≈ kan2(2Δ)1/ 2 (3)

and

d(Vb)dV

=d

dVbkan2 (2Δ)1 / 2[ ]= an2(2Δ)1/ 2 d

dV(bk) (4)

Show that

dτdλ

= −Ln2Δcλ

Vd 2(Vb)

dV 2 (5)

and that the waveguide dispersion coefficient is

Dw =dτLdλ

= −n2Δcλ

Vd 2(Vb)

dV 2 (6)

Figure 2Q12 shows the dependence of V[d2(Vb)/dV2] on the V-number. In the range 1.5 < V < 2.4,

Vd2 (Vb)

dV2 ≈1.984

V2

Show that,

Dw ≈ −n2Δcλ

1.984V2 = −

(n1 − n2 )cλ

1.984V 2 (7)

which simplifies to

Dw ≈ −1.984

c(2πa)22n2

λ (8)

Equation (2) should really have Ng2 instead of n2 in which case Eq. (8) would be

Dw ≈ −

1.984Ng 2

c(2πa)22n22 λ (9)

Consider a fiber with a core of diameter of 8 μm and refractive index of 1.468 and a cladding refractive index of 1.464 both refractive indices at 1300 nm. Suppose that a 1.3 μm laser diode with a spectral linewidth of 2 nm is used to provide the input light pulses. Estimate the waveguide dispersion per kilometer of fiber using Eqs. (6) and (9).

0

0.5

1

1.5

0 1 2 3V - number

V[d2(Vb)/dV2]

[d2(Vb)/dV2] vs. V-number for a step index fiber. Figure 2Q12

Solution Waveguide dispersion arises as a result of the dependence of the propagation constant on the V-number which depends on the wavelength. It is present even when the refractive index is constant; no material dispersion. Let us suppose that n1 and n2 are wavelength (or k) independent. Suppose that β is the propagation constant of mode lm and k = 2π/λ where λ is the free space wavelength. Then the normalized propagation constant b is defined as,

b =(β / k)2 − n2

2

n12 − n2

2 (1)

Show that for small normalized index difference Δ = (n1 − n2)/n1, Eq. (1) approximates to

b =(β / k) − n2

n1 − n2

(2)

which gives β as,

β = n2k[1 + Δb] (3)

The group velocity is defined and given by

vg =

dωdβ

= cdkdβ

Thus, the propagation time τ of the mode is

τ =

Lvg

=Lc

dβdk

⎛ ⎝

⎞ ⎠ =

Ln2

c+

Ln2Δc

d(kb)dk

(4)

where we assumed Δ ≈ constant (does not depend on the wavelength). Given the definition of V,

V = ka[n12 − n2

2 ]1 / 2 = ka[(n1 + n2 )(n1 − n2 )]1/ 2

≈ ka[(n1 + n2 )n1n1 − n2

n1

⎛

⎝ ⎜ ⎞

⎠ ⎟ ]1/ 2

≈ ka[2n2n1Δ]1/ 2 ≈ kan2 (2Δ)1/ 2

(5)

From Eq. (5),

d(Vb)dV

=d

dVbkan2 (2Δ)1 / 2[ ]= an2(2Δ)1/ 2 d

dV(bk)

This means that τ depends on V as,

τ =Ln2

c+

Ln2 Δc

d(Vb)dV

(6)

Dispersion, that is, spread δτ in τ due to a spread δλ can be found by differentiating Eq. (6) to obtain,

dτdλ

=Ln2 Δ

cdVdλ

ddV

d(Vb)dV

=Ln2Δ

c−

Vλ

⎛ ⎝

⎞ ⎠

d2(Vb)dV 2

= − Ln2 Δcλ

V d2 (Vb)dV 2

(7)

The waveguide dispersion coefficient is defined as

Dw =dτLdλ

= −n2Δcλ

Vd 2(Vb)

dV 2 (8)

Figure 2Q5 shows the dependence of V[d2(Vb)/dV2] on the V-number.

In the range 2 < V < 2.4,

Vd2 (Vb)

dV2 ≈1.984

V2

so that Eq. (8) becomes,

Dw ≈ −n2Δcλ

1.984V2 = −

(n1 − n2 )cλ

1.984V 2 (9)

We can simplify this further by using

Dw ≈ −n2Δcλ

1.984V2 ≈ −

1.984n2 Δcλ

λ2πan2 (2Δ)1 / 2

⎡

⎣ ⎢ ⎤

⎦ ⎥

1/ 2

Dw ≈ −1.984

c(2πa)22n2

λ (10)

Equation (6) should really have Ng2 instead of n2 in which case Eq. (10) would be

Dw ≈ −

1.984Ng 2

c(2πa)22n22 λ (11)

Consider a fiber with a core of diameter of 8 μm and refractive index of 1.468 and a cladding refractive index of 1.464 both refractive indices at 1300 nm. Suppose that a1.3 μm laser diode with a spectral linewidth of 2 nm is used to provide the input light pulses. Estimate the waveguide dispersion per kilometer of fiber using Eqs. (8) and (11).

V =2πa

λn1

2 − n12( )1/ 2

=2π(4 μm) 1.4682 −1.4642( )1/ 2

(1.3 μm) = 2.094

and Δ = (n1 − n2)/n1 = (1.468-1.464)/1.468 = 0.00273.

From the graph, Vd2(Vb)/dV2 = 0.45,

Dw = −n2 Δcλ

Vd2 (Vb)

dV 2 = −(1.464)(2.73 ×10−3 )

(3 ×108 m s-1)(1300 ×10−9 m)(0.45)

∴ Dw ≈ - 4.6×10-6 s m-2 or -4.6 ps km-1 nm-1

Using Eq. (10)

Dw ≈ −1.984

c(2πa)22n2

λ = −1.984(1300 ×10−9 m)

(3 ×108 m s-1 )[2π × 4 ×10−6 m]2 2(1.464)]

∴ Dw ≈ - 4.6×10-6 s m-2 or -4.6 ps km-1 nm-1

For Δλ1/2 = 2 nm we have,

Δτ1/2 = |Dw|L Δλ1/2 = (4.6 ps km-1 nm-1)(2 nm) = 9.2 ps/km

2.13 Profile dispersion Total dispersion in a single mode step index fiber is primarily due to material dispersion and waveguide dispersion. However, there is an additional dispersion mechanism called profile dispersion that arises from the propagation constant β of the fundamental mode also depending on the refractive index difference Δ. Consider a light source with a range of wavelengths δλ coupled into a step index fiber. We can view this as a change δλ in the input wavelength λ. Suppose that n1, n 2, hence Δ

depends on the wavelength λ. The propagation time, or the group delay time, τg per unit length is

τg =

1vg

=1c

dβdk

⎛ ⎝

⎞ ⎠ (1)

δτg

δλ=

∂τ g

∂n1

∂n1

∂λ+

∂τ g

∂V∂V∂λ

+∂τ g

∂Δ∂Δ∂λ

(2)

Total dispersion = Materials dispersion (due to ∂n1/∂λ)

+ Waveguide dispersion (due to ∂V/∂λ)

+ Profile dispersion (due to ∂Δ/∂λ)

where the last term is due to Δ depending on λ; although small this is not zero. Even the above statement in Eq. (2) is over simplified but nonetheless provides an insight into the problem. The total intramode (chromatic) dispersion coefficient Dch is then given by

Dch = Dm + Dw + Dp (3)

where Dm, Dw, Dp are material, waveguide and profile dispersion coefficients respectively. The waveguide dispersion is given by Eq. (8) in Question 2.6 and the profile dispersion coefficient is (very) approximately,

Dp ≈ −

Ng1

cV

d 2(Vb)dV2

⎛ ⎝ ⎜ ⎞

⎠ dΔdλ

⎛ ⎝

⎞ ⎠ (4)

where b is the normalized propagation constant and Vd2(Vb)/dV2 vs. V is shown in Figure 2Q12. The term Vd2(Vb)/dV2 ≈ 1.984/V2.

Consider a fiber with a core of diameter of 8 μm. The refractive and group indices of the core and cladding at λ = 1.55 μm are n1 = 1.4504, n 2 = 1.4450, Ng1 = 1.4676, Ng 2 = 1.4625, and dΔ/dλ = 161 m-1. Estimate the waveguide and profile dispersion per km of fiber per nm of input light linewidth at this wavelength.

Solution Total dispersion in a single mode step index fiber is primarily due to material dispersion and waveguide dispersion. However, there is an additional dispersion mechanism called profile dispersion that arises from the propagation constant β of the fundamental mode also depending on the refractive index difference Δ. Consider a light source with a range of wavelengths δλ coupled into a step index fiber. We can view this as a change δλ in the input wavelength λ. Suppose that n1, n 2, hence Δ depends on the wavelength λ. The propagation time, or the group delay time, τg per unit length is

τg =

1vg

=1c

dβdk

⎛ ⎝

⎞ ⎠ (1)

δτg

δλ=

∂τ g

∂n1

∂n1

∂λ+

∂τ g

∂V∂V∂λ

+∂τ g

∂Δ∂Δ∂λ

(2)

Total dispersion = Materials dispersion (due to ∂n1/∂λ)

+ Waveguide dispersion (due to ∂V/∂λ)

+ Profile dispersion (due to ∂Δ/∂λ)

where the last term is due Δ depending on λ; although small this is not zero. Even the above statement in Eq. (2) is over simplified but nonetheless provides an sight into the problem. The total intramode (chromatic) dispersion coefficient Dch is then given by

Dch = Dm + Dw + Dp (3)

where Dm, Dw, Dp are material, waveguide and profile dispersion coefficients respectively. The waveguide dispersion is given by Eq. (8) in Question 2.6 and the profile dispersion coefficient away is (very) approximately,

Dp ≈ −

Ng1

cV

d 2(Vb)dV2

⎛ ⎝ ⎜ ⎞

⎠ dΔdλ

⎛ ⎝

⎞ ⎠ (4)

where b is the normalized propagation constant and Vd2(Vb)/dV2 vs. V is shown in Figure 2Q6. The term Vd2(Vb)/dV2 ≈ 1.984/V2.

Consider a fiber with a core of diameter of 8 μm. The refractive and group indexes of the core and cladding at λ = 1.55 μm are n1 = 1.4504, n 2 = 1.4450, Ng1 = 1.4676, Ng 2 = 1.4625. dΔ/dλ = 161 m-1.

V =2πaλ

n12 − n1

2( )1/ 2=

2π(4 μm) 1.45042 −1.44502( )1 / 2

(1.55 μm) = 2.03

and Δ = (n1 − n2)/n1 = (1.4504-1.4450)/1.4504 = 0.00372

From the graph in Figure 2Q12, Vd2(Vb)/dV2 = 0.5,

Profile dispersion:

Dp = −

Ng1

cV

d2 (Vb)dV2

⎛ ⎝ ⎜ ⎞

⎠ dΔdλ

⎛ ⎝

⎞ ⎠ = −

1.46763 ×108 m s−1

0.5( )161 m−1( )

∴ Dp = 3.8 × 10-7 s m-1 m-1 or 0.38 ps km-1 nm-1

Waveguide dispersion:

Dw ≈ −1.984

c(2πa)22n2

λ = −1.984(1500 ×10−9 m)

(3 ×108 m s-1 )[2π × 4 ×10−6 m]2 2(1.4450)]

∴ Dw ≈ − 5.6 ps km-1 nm-1

Profile dispersion is more than 10 times smaller than waveguide dispersion.

3.5 AlGaAs LED emitter An AlGaAs LED emitter for use in a local optical fiber network has the output spectrum shown in Figure 3Q5. It is designed for peak emission at 820 nm at 25°C.

a What is the linewidth Δλ between half power points at temperatures −40°C, 25°C and 85°C? What is the empirical relationship between Δλ and T given three temperatures and how does this compare with Δ(hυ) ≈ 2.5kBT − 3kB BBT?

b Why does the peak emission wavelength increase with temperature?

c Why does the peak intensity decrease with temperature?

d What is the bandgap of AlGaAs in this LED?

e The bandgap, Eg, of the ternary alloys AlxGa1-xAs follows the empirical expression,

Eg(eV) = 1.424 + 1.266x + 0.266x2.

What is the composition of the AlGaAs in this LED? f When the forward current is 40 mA, the voltage across the LED is 1.5V and the optical power that is coupled into a multimode fiber through a lens is 25 μW. What is the efficiency?

800 900

–40°C

25°C

85°C

0

1

740

Relative spectral output power

840 880Wavelength (nm)

The output spectrum from GaAlAs LED. Valuesnormalized to peak emission at 25°C.

Figure 3Q5

Solution a We note that the emitted wavelength is related to the photon energy Eph by

λ = c/υ = hc/Eph.

If we differentiate λ with respect to photon energy Eph we get

dλdE ph

= −hc

Eph2

We can represent small changes or intervals (or Δ) by differentials, e.g. Δλ/ΔEph ≈ |dλ/dEph|, then

Δλ ≈hc

Eph2 ΔEph

We are given the energy width of the output spectrum, ΔEph = Δ(hυ) ≈ 3kBT. Then, using the latter and substituting for E

B

ph in terms of λ we find,

Δλ ≈ λ2 3kBThc

or Δλ ≈ λ2 3kBThc

Temperature →

Parameter ↓

−40 °C 25 °C 85 °C Comment

λpeak (nm) 804 820 837

Δλ (nm) (Measured) 30 40 48

Δλ (nm) (Calculated); ΔEph = 2.5 kT 26.2 34.8 43.6

Δλ (nm) (Calculated); ΔEph = 3 kT 31.4 41.7 52.3 Very close

B

B

B

0

10000

20000

30000

40000

50000

60000

70000

0 100 200 300 400

Δλ/(λ2)

Temperature (K)

Best line forced through zero is

Δλ/(λ2) = 1956T ; R2 = 0.9932374

The theory predicts that Δλ/λ2 vs T should be a straight line because,

Δλ ≈ λ2 ΔEph

hc

so that Δλλ2 =

mkBThc

where ΔEph = mkBT and m is a numerical constant that represents the ratio Δ(hυ)/(kB BBT) and is determined from the slope of the Δλ/λ2 vs. T plot.

The three points plotted in the figure seems to follow this behavior. The best line forced through zero has a slope that indicates m = 2.8.

b The bandgap decreases with temperature

c There are two factors to consider. (i) Spectral intensity means intensity per unit wavelength, that is, dI/dλ. The integration of the spectral curve gives the total intensity, the total number of photons emitted per unit area per unit time. As the spectrum broadens with temperature we would naturally expect the peak to decrease with temperature. (ii) Higher the temperature, the stronger are the lattice vibrations (there are more phonons).

Indirect or radiationless transitions, those that do not emit photons, require phonons (lattice vibrations) which encourage indirect transitions. Thus increasing the temperature increases indirect transitions at the expense of direct transitions and the light intensity decreases. If (ii) was totally absent then the areas under the curves for all the three spectra would be identical.

d Use the peak emission wavelength to find Eg as follows:

At -40 °C (233 K) , λpeak = 804 nm.

At 25 °C (298 K), λpeak = 820 nm.

At 85 °C (358 K), λpeak = 837 nm.

We first note that we need the required bandgap Eg at the wavelength of interest. The photon energy at peak emission is hc/λpeak = Eg + kBT. Then, B

Eg =ch

eλpeak

−kBTe

and at λpeak = 820×109 m, taking T = 25 + 273K,

Eg =(3×108 )(6.626 ×10−34)(1.6 ×10−19)(820 ×10−9)

− 0.0257 eV = 1.4863 eV

e The bandgap Eg of the ternary alloys AlxGa1-xAs follows the empirical expression,

Eg(eV) = 1.424+1.266x+0.266x2.

∴ Eg(eV) = 1.4863 = 1.424 + 1.266x + 0.266x2.

Solving for x we find, x = 0.05.

f From the definition of efficiency η,

η = Output optical powerInput electrical power

=Po

IV=

25 ×10−6 W(40 ×10−3 A)(1.5 V)

= 0.000417 = 0.0417%

3.6 III-V compound semiconductors in optoelectronics Figure 3Q6 represents the bandgap Eg and the lattice parameter a in the quarternary III-V alloy system. A line joining two points represents the changes in Eg and a with composition in a ternary alloy composed of the compounds at the ends of that line. For example, starting at GaAs point, Eg = 1.42 eV and a = 0.565 nm, and Eg decreases and a increases as GaAs is alloyed with InAs and we move along the line joining GaAs to InAs. Eventually at InAs, Eg = 0.35 eV and a = 0.606 nm. Point X in Figure 3Q6 is composed of InAs and GaAs and it is the ternary alloy InxGa1-xAs. It has Eg = 0.7 eV and a = 0.587 nm which is the same a as that for InP. InxGa1−xAs at X is therefore lattice matched to InP and can hence be grown on an InP substrate without creating defects at the interface.

Further, InxGa1-xAs at X can be alloyed with InP to obtain a quarternary alloy InxGa1-xAsyP1-y whose properties lie on the line joining X and InP and therefore all have

the same lattice parameter as InP but different bandgap. Layers of InxGa1-xAsyP1-y with composition between X and InP can be grown epitaxially on an InP substrate by various techniques such as liquid phase epitaxy (LPE) or molecular beam expitaxy (MBE) .

The shaded area between the solid lines represents the possible values of Eg and a for the quarternary III-V alloy system in which the bandgap is direct and hence suitable for direct recombination.

The compositions of the quarternary alloy lattice matched to InP follow the line from X to InP.

a Given that the InxGa1-xAs at X is In0.535Ga0.465As show that quarternary alloys InxGa1-xAsyP1-y are lattice matched to InP when y = 2.15x.

b The bandgap energy Eg, in eV for InxGa1-xAsyP1-y lattice matched to InP is given by the empirical relation,

Eg (eV) = 1.35 − 0.72y + 0.12 y2

Find the composition of the quarternary alloy suitable for an emitter operating at 1.55 μm.

0.20.40.60.8

11.21.41.61.8

22.22.42.6

0.54 0.55 0.56 0.57 0.58 0.59 0.6 0.61 0.62Lattice constant, a (nm)

GaP

GaAs

InAs

InP

Direct bandgapIndirect bandgap

In0.535Ga0.465AsX

Quaternary alloyswith direct bandgap

In1-xGaxAs

Quaternary alloyswith indirect bandgap

Eg (eV)

Bandgap energy Eg and lattice constant a for various III-V alloys of GaP, GaAs,InP and InAs. A line represents a ternary alloy formed with compounds from theend points of the line. Solid lines are for direct bandgap alloys whereas dashedlines for indirect bandgap alloys. Regions between lines represent quaternaryalloys. The line from X to InP represents quaternary alloys In1-xGaxAs1-yPy madefrom In0.535Ga0.465As and InP which are lattice matched to InP.

Figure 3Q6

Solution

a The III−V quaternary alloy is made of (In0.535Ga0.465As) alloyed with InP. Suppose that z fraction of In0.535Ga0.465As is alloyed with (1 − z) fraction of InP.

z = 0 for InP and z = 1 for In0.535Ga0.465As

The formula for the quaternary alloy is In1−xGaxAsyP1−y.

Thus, z(In0.535Ga0.465As) + (1−z)InP = In1−xGaxAsyP1−y

Use x = x′ to represent point X or (In0.535Ga0.465As), that is, x′ = 0.535 for the In content.

We can now balance the contents for each element.

For P 1 − z = 1 − y ∴ z = y

For Ga (1 − x′)z = x ∴ y = 2.15x

For In x′z + (1 − z) = 1 − x ∴ z = x/(1 − x')

For As z = y ∴ y = 2.15x

b Consider a lattice matched InGaAsP to InP. Given λ = 1.55 μm.

Eg = 1.24/λ = 0.80 eV

Thus, Eg (eV) = 1.35 − 0.72y + 0.12 y2 = 0.80 eV

∴ 0.12y2 − 0.72y + 0.55 = 0

∴ y = 0.90, x = 0.42

∴ In0.58Ga0.42As0.9P0.1

3.7 External conversion efficiency The external power or conversion efficiency ηext is defined as

ηext =Optical power outputElectrical power input

=Po

IV

One of the major factors reducing the external power efficiency is the loss of photons in extracting the emitted photons which suffer reabsorption in the pn junction materials, absorption outside the semiconductors and various reflections at interfaces.

The total light output power from a particular AlGaAs red LED is 2.5 mW when the current is 50 mA and the voltage is 1.6 V. Calculate its external conversion efficiency.

Solution

ηext =Po

IV=

2.5 ×10−3 W(50 ×10−3 A)(1.6 V)

= 0.03125 = 3.125 %

3.9 SLEDs and ELEDs Experiments carried out on an AlGaAs SLED (surface emitting LED) and an ELED (edge emitting LED) give the light output power vs. current data in Table 3Q9.

a Show that the output light power vs. current characteristics are not linear.

b By plotting the optical power output (Po) vs current (I) data on a log-log plot show that Po ∝ In . Find n for each LED.

Table 3Q9

Light output power vs. DC current for surface and edge emitting LEDs. SLED I (mA) 25 50 75 100 150 200 250 300

SLED Light output power (mW) Po

1.04 2.07 3.1 4.06 5.8 7.6 9.0 10.2

ELED I (mA) 25 50 75 100 150 200 250 300

ELED Light output power (mW) Po

0.46 0.88 1.28 1.66 2.32 2.87 3.39 3.84

Solution

♦

♦

♦♦

♦♦

♦♦

Ο

ΟΟ

ΟΟ

ΟΟΟ

0.1

1

10

20

10 100 1000

♦ SLED

Ο ELED

Current (mA)

Light Output Power (mW)

n = 0.86

n = 0.92

SLED Po = (0.00556)I0.925

Correlation coefficient = 0.9979

ELED Po = (0.0306)I0.857

Correlation coefficient = 0.9971 3.11 Internal quantum efficiency The internal efficiency ηint gauges what fraction of electron hole recombinations in the forward biased pn junction are radiative and therefore lead to photon emission. Nonradiative transitions are those in which an electron and a hole recombine through a recombination center such as a crystal defect or an impurity and emit phonons (lattice vibrations). By definition,

ηint =Rate of radiative recombination

Total rate of recombination (radiative and nonradiative) (1)

or ηint =

1τ r

1τr

+ 1τ nr

(2)

where τr is the mean lifetime of a minority carrier before it recombines radiatively and τnr is the mean lifetime before it recombines via a recombination center without emitting a photon. The total current I is determined by the total rate of recombinations whereas the number of photons emitted per second (Φph) is determined by the rate of radiative recombinations.

ηint =Photons emitted per second

Total carriers lost per second=

Φ ph

I / e=

Pop(int) / hυI / e

(3)

where Pop(int) is the optical power generated internally (not yet extracted).

For a particular AlGaAs LED emitting at 850 nm it is found that τr = 50 ns and τnr = 100 ns. What is the internal optical power generated at a current of 100 mA?

Solution

Consider ηint =

1τ r

1τr

+ 1τ nr

=

150 ns

150 ns

+ 1100 ns

= 0.667 = 66.7 %

From ηint =Pop(int) / hυ

I / e

Pop(int) = η int(Ie)hυ = ηint

Ihceλ

= (0.667)(100 ×10−3)(6.626 ×10−34 )(3 ×108 )

(1.6 ×10−19)(850 ×10−9 )

= 0.097 W or 97 mW